Question

Bag $ I $ contains $ 3 $ red and $ 4 $ black balls and Bag $ II $ contains $ 4 $ red and $ 5 $ black. One ball is transferred from bag $ I $ to bag $ II $ and then a ball is drawn from bag $ II $ . The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Answer 1

Hint: First find the number of balls left in each bag after the transfer is done. Then use Bayes’ Theorem on the conditional probability to solve this question.

Complete step-by-step answer:
Let’s consider the events $ {E_1} $ and $ {E_2} $ as follows.
 $ {E_1} = $ ball transferred from a bag $ I $ to bag $ II $ is red.
 $ {E_2} = $ Ball transferred from bag $ I $ to bag $ II $ is black.
 $ A = $ Ball drawn from bag $ II $ is red
 $ P({E_1}) $ is the probability of transferring red ball from bag $ I $ to bag $ II $
 $ \Rightarrow P({E_1}) = \dfrac{3}{7} $ (since, bag $ I $ contains $ 3 $ red balls and $ 4 $ black balls)
 $ P({E_2}) $ is the probability of transferring black ball from bag $ I $ to bag $ II $
 $ \Rightarrow P({E_2}) = \dfrac{4}{7} $ (since, Bag $ I $ contains $ 3 $ red and $ 4 $ black balls)
Red balls are added to bag $ II $ .
Therefore, the total number of balls in bag $ II $ is $ 10 $
And number of red balls is $ 4 + 1 = 5 $
By using the formula of conditional probability, we can write
Probability of drawing red ball from bag $ II $ when red ball is already transferred from bag $ I $ to bag $ II $ is given by $ P(A|{E_1}) $
 $ \Rightarrow P(A|{E_1}) = \dfrac{5}{{10}} $
 $ \Rightarrow P(A|{E_1}) = \dfrac{1}{2} $
Now, let is assume that a black ball is added to bag $ II $
Therefore, the total number of balls in bag $ II $ is 10
And the number of red balls is 4
Then, probability of drawing red ball from bag $ II $ when black ball is already transferred from bag $ I $ to bag $ II $ is given by $ P(A|{E_2}) $
 $ \Rightarrow P(A|{E_1}) = \dfrac{4}{{10}} $
 $ \Rightarrow P(A|{E_1}) = \dfrac{2}{5} $
Probability of transferring a black ball when the ball drawn from bag $ II $ is found to be red is given by $ P({E_2}|A) $
Using Bayes’ theorem, we can write
\[P({E_2}|A) = \dfrac{{P({E_2})P(A|{E_2})}}{{P({E_1})P(A|{E_1}) + P({E_2})P(A|{E_2})}}\]
 $ = \dfrac{{\dfrac{4}{7} \times \dfrac{2}{5}}}{{\dfrac{3}{7} \times \dfrac{1}{2} + \dfrac{4}{7} \times \dfrac{2}{5}}} $
Simplifying it, we get
\[P({E_2}|A) = \dfrac{{\dfrac{8}{{35}}}}{{\dfrac{1}{7}\left( {\dfrac{3}{2} + \dfrac{8}{5}} \right)}}\]
\[ = \dfrac{{\dfrac{8}{{35}} \times 7}}{{\dfrac{{15 + 16}}{{10}}}}\]
 $ = \dfrac{{\dfrac{8}{5}}}{{\dfrac{{31}}{{10}}}} $
 $ = \dfrac{8}{5} \times \dfrac{{10}}{{31}} $ $ \left( {\because \dfrac{{\dfrac{a}{b}}}{{\dfrac{p}{q}}} = \dfrac{a}{b} \times \dfrac{q}{p}} \right) $
 $ = \dfrac{{8 \times 2}}{{31}} $
 $ \Rightarrow P({E_2}|A) = \dfrac{{16}}{{31}} $
Therefore, probability of transferring a black ball from bag $ I $ to bag $ II $ when the ball drawn from bag $ II $ is found to be red is given by $ P({E_2}|A) $

Note: Conditional probability is tricky to understand. $ P(A|B) $ is the probability of occurrence of event A when event B has already happened or true. $ P(B|A) $ is the probability of occurrence of event B when event A has already happened or true.
 $ P(A|B) \ne P(B|A) $ So, be careful while using conditional probability.